Bernd Finkbeiner Date: October 25, Automata, Games, and Verification: Lecture 2

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1 Bernd Finkeiner Dte: Octoer 25, 2012 Automt, Gmes, nd Verifiction: Lecture 2 2 Büchi Automt Definition1 AnondeterministicBüchiutomtonAoverlphetΣistuple(S,I,T,F): S : finitesetof sttes I S:susetof initilsttes T S Σ S : setof trnsitions F S : susetof cceptingsttes Now we define how Büchi utomton uses n infinite word s input. Notice tht we do not refer to cceptnce in this definition. Definition2 A run of nondeterministic Büchi utomtonaon n infinite input word α = σ 0 σ 1 σ 2...isninfinitesequenceofsttess 0,s 1,s 2,...suchthtthefollowinghold: s 0 I forll i ω,(s i,σ i,s i+1 ) T Exmple: A D B C In the utomton shown the set of sttes re S={A,B,C,D}, the initil set of sttes re I= {A} (indicted with pointing rrow with no source), the trnsitions T={(A,,B),(B,,C), (C,,D),(D,,A)}rethereminingrrowsinthedigrm,ndthesetofcceptingsttesis F ={D}(doule-lined stte circle). Oninput...theBüchiutomtonshownhsonlytherun: ABCDABCDABCD...

2 Determinism is property of mchines tht cn only rect in unique wy to their input. The following definition mkes this cler for Büchi utomt. Definition 3 A Büchi utomton A is deterministic when T is prtil function(with respect to the next input letter nd the current stte): σ Σ, s,s 0,s 1 S.(s,σ,s 0 ) T nd(s,σ,s 1 ) T s 0 = s 1 nd I is singleton. (By Büchi utomton we usully men nondeterministic Büchi utomton.) Definition4 Theinfinitysetofninfiniteword α Υ ω oversomelphet Υ isthesetin(α)= {υ Υ i j. j i ndα(j)=υ} Definition5 ABüchiutomtonAcceptsninfinitewordαif: thereisrunr=s 0 s 1 s 2...ofαonA r isccepting: In(r) F The lnguge recognized y Büchi utomton A is defined s follows: L(A)={α Σ ω Acceptsα} Exmple: The utomton from the previous exmple hs the lnguge{...}. Comment: A deterministic Büchi utomtona=(s,i,t,f) defines prtil function 1 from Σ ω tosetofrunsr S ω. Definition6 An ω-lnguge L is Büchi recognizle if there is Büchi utomtonasuch tht L(A)=L. Exmple:Thesingletonω-lngugeL={σ}withσ=...isnotBüchirecognizle. (Note tht ll finite lnguges of finite words re NFA-recognizle. Anlog result does not hold for Büchi-utomt) SupposethereisBüchiutomtonAwithL(A)=L. Letr=s 0 s 1...encceptingrunonσ. SinceF isfinite,thereexists k,k ωwith k<k nds k = s k F. r = r 0...r k 1(r k...r k 1) ω isncceptingrunon σ = σ(0)...σ(k 1)(σ(k)...σ(k 1)) ω. Hence,σ L(A). Contrdiction. Definition 7 A Büchi utomton is complete if its trnsition reltion contins function: s S,σ Σ. s S.(s,σ,s ) T 1 Aprtilfunctionisfunctionthtisnotdefinedonlloftheelementsofitsdomin.

3 Theorem1 For every Büchi utomtona, there is complete Büchi utomtona such tht L(A)=L(A ). WedefineA intermsofthecomponentss,i,t,f ofa: S = S {f} f new I = I T = T {(s,σ, f) / s.(s,σ,s ) T} {(f,σ, f) σ Σ} F = F The runs ofa re superset of those ofasince we hve dded sttes nd trnsitions. Furthermore, on ny infinite input word α the ccepting runs ofanda correspond, ecusenyrunthtreches f stysin f,ndsince f/ F,suchrunisnotccepting. Exmple: Completing the utomton from the previous exmples we otin the following utomton: A B D f C, Unless we specify otherwise, we will only consider complete utomt when we prove results. Comment:AcompletedeterministicBüchiutomtonA=(S,I,T,F)myeviewedstotl functionfrom Σ ω to S ω. Acomplete(possilynondeterministic)Büchiutomtonproducest lestonerunforeveryσ ω inputword. 3 ω-regulr Lnguges Definition 8 The ω-regulr expressions re defined s follows. IfRisnregulrexpressionwhereε/ L(R), thenr ω isnω-regulrexpression. L(R ω )=L(R) ω wherel ω ={u 0 u 1... u i L, u i >0forll i ω}forl Σ.

4 IfRisregulrexpressionndU isnω-regulrexpression, thenr U isnω-regulrexpression. L(R U)=L(R) L(U) wherel 1 L 2 ={r u r L 1,u L 2 }forl 1 Σ,L 2 Σ ω. IfU 1 ndu 2 reω-regulrexpressions, thenu 1 +U 2 isnω-regulrexpression. L(U 1 +U 2 )=L(U 1 ) L(U 2 ). Definition9 An ω-regulr lnguge is finite union of ω-lnguges of the form U V ω where U,V Σ reregulrlnguges. Theorem2 IfL 1 ndl 2 rebüchirecognizle,thensoisl 1 L 2. LetA 1 nda 2 ebüchiutomtthtrecognizel 1 nd L 2,respectively. Weconstructn utomtona forl 1 L 2 : S = S 1 S 2 (w.l.o.g. wessumes 1 S 2 = ); I = I 1 I 2 ; T = T 1 T 2 ; F = F 1 F 2. L(A ) L(A 1 ) L(A 2 ): For α L(A ), we hve n ccepting run r=s 0 s 1 s 2... of α in A. Ifs 0 S 1,thenr isncceptingrunona 1,otherwises 0 S 2 ndr isncceptingrun ona 2. L(A ) L(A 1 ) L(A 2 ): For i {1,2} nd α L(A i ), there is n ccepting run r= s 0 s 1 s 2...onA i. Therunr iscceptingforαina. Theorem3 IfL 1 ndl 2 rebüchirecognizle,thensoisl 1 L 2. WeconstructnutomtonA froma 1 nda 2 : S = S 1 S 2 {1,2} I = I 1 I 2 {1} T ={((s 1,s 2,1),σ,(s 1,s 2,1)) (s 1,σ,s 1 ) T 1,(s 2,σ,s 2 ) T 2,s 1 / F 1 } {((s 1,s 2,1),σ,(s 1,s 2,2)) (s 1,σ,s 1 ) T 1,(s 2,σ,s 2 ) T 2,s 1 F 1 } {((s 1,s 2,2),σ,(s 1,s 2,2)) (s 1,σ,s 1 ) T 1,(s 2,σ,s 2 ) T 2,s 2 / F 2 } {((s 1,s 2,2),σ,(s 1,s 2,1)) (s 1,σ,s 1 ) T 1,(s 2,σ,s 2 ) T 2,s 2 F 2 } F ={(s 1,s 2,2) s 1 S 1,s 2 F 2 } L(A )=L(A 1 ) L(A 2 ): r =(s1 0,s0 2,t0 )(s1 1,s1 2,t1 )... isrunofa oninputword σ iff r 1 = s1 0s isrunof A 1 onσ ndr 2 = s2 0s1 2...isrunofA 2 onσ.

5 r iscceptingiffr 1 iscceptingndr 2 isccepting. Theorem4 IfL 1 isregulrlngugendl 2 isbüchirecognizle,thenl 1 L 2 isbüchi-recognizle. LetA 1 e finite-word utomton tht recognizes L 1 nda 2 e Büchi utomton tht recognizesl 2. Weconstruct: S = S 1 S 2 (w.l.o.g. wessumes 1 S 2 = ); I ={ I 1 ifi 1 F 1 = I 1 I 2 otherwise; T = T 1 T 2 {(s,σ,s ) (s,σ, f) T 1, f F 1,s I 2 }; F = F 2. Theorem5 IfLisregulrlngugethenL ω isbüchirecognizle. LetAefinitewordutomton;letw.l.o.g. ε/ L(A). Step 1: Ensure tht ll initil sttes hve no incoming trnsitions. We modifyas follows: S = S {s new }; I ={s new }; T = T {(s new,σ,s ) (s,σ,s ) T forsomes I}; F = F. This modifiction does not ffect the lnguge of A. Step2: Addloop: S = S ;I = I ; T = T {(s,σ,s new (s,σ,s ) T nds F }; F = I. L(A ) L(A ) ω : Assumeα L(A )nds 0 s 1 s 2...isncceptingrunforαinA. Hence,s i = s new F = I forinfinitelymnyindices i: i 0,i 1,i 2,... ThisprovidesseriesofrunsinA : runs 0 s 1...s i1 1s onw 1 = α(0)α(1)...α(i 1 1)forsomes F ; runs i1 s i s i2 1s onw 2 = α(i 1 )α(i 1 +1)...α(i 2 1)forsomes F ;

6 ... Thisyieldsw k L(A )forevery k 1. Hence,α L(A ) ω. L(A ) L(A ) ω : Letα=w 1 w 2 w 3 Σ ω suchthtw k L(A )forll k 1. Forech k,wechoosencceptingruns0 ksk 1 sk 2...sk n k ofa onw k. Hence,s0 k I ndsn k k F forll k 1. Thus, s0...s 1 1 n 1 1s0...s 2 2 n 2 1s0...s 3 3 n isncceptingrunonαina. Hence,α L(A ). Theorem 6(Büchi s Chrcteriztion Theorem(1962)) An ω-lnguge is Büchi recognizle iff it is ω-regulr.